On the general equation of the second degree

Transcription

1 On the general equation of the second degree S Kesavan The Institute of Mathematical Sciences, CIT Campus, Taramani, Chennai Abstract We give a unified treatment of the general equation of the second degree in two real variables in terms of the eigenvalues of the matrix associated to the quadratic terms and describe the solution sets in all cases 1

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3 1 Introduction The study of the general equation of the second degree in two variables used to be a major chapter in a course on analytic geometry in the undergraduate mathematics curriculum for a long time The equation usually represents a pair of straight lines or a conic In the latter case the method of tracing a conic was to compute the trigonometric ratios of the angle that the axes of the conic make with the coordinate axes and then rotate the coordinate axes to reduce the equation to the normal form These computations could be tedious Further in most classical text books the treatment is rather incomplete and the cases when the solution set is degenerate (especially when it contains a single point or is empty) are not carefully explained The aim of this note is to study all cases of the equation in a unified manner By just computing the eigenvalues and eigenvectors of the 2 2 real symmetric matrix associated to the quadratic terms, we can just read off the properties of the solution set and also write down the equations of various features of the set very easily This approach neatly brings out some of the connections between linear algebra and geometry 2 Some linear algebra Consider the following 2 2 real symmetric matrix: [ ] a h A = h b Its characteristic equation is The discriminant is λ 2 (a + b)λ + (ab h 2 ) = 0 (a + b) 2 4ab + 4h 2 = (a b) 2 + 4h 2 0 and so both its eigenvalues are real and are given by (a + b) ± (a b) 2 + 4h 2 2 3

5 eigenvalues of A (in the same order corresponding to the column vectors of P ), then A = P DP T The same result is true if A is a complex hermitian (ie self-adjoint) matrix, in which case we replace P T in the preceding relation by P, the conjugate transpose of P (and P will be a unitary matrix) Example 21 Let A = [ Then, its characteristic equation is λ 2 10λ + 16 = 0 and its eigenvalues are, therefore, λ 1 = 2 and λ 2 = 8 Corresponding to λ 1 = 2, we get the equation 5x 3y = 2x or, equivalently, x = y Thus the eigenvectors corresponding to λ 1 are scalar multiples of (1,1) Normalizing, we get (u 1, u 2 ) = ( 1 2, ] ) 1 2 Corresponding to the eigenvalue λ 2 = 8, we get the equation 5x 3y = 8x, or, equivalently, x = y Thus all eigenvectors corresponding to this eigenvalue are scalar multiples of (1, 1) and, normalizing, we get ( 1 (v 1, v 2 ) = 2, 1 ) 2 Thus P = [ ] and it is easy to verify that A = P DP T where [ ] 2 0 D = The homogeneous equation Consider the following homogeneous equation of the second degree in two real variables: ax 2 + 2hxy + by 2 = 0 (31) Let us denote by S the set of all points (x, y) in the plane which satisfy this equation Our aim is to determine this set 5

8 4 The inhomogeneous equation Let us now consider the inhomogeneous equation ax 2 + 2hxy + by 2 = 1 (41) With the notations established in the preceding sections, this equation reduces to λ 1 x 2 + λ 2 y 2 = 1 (42) If S is the solution set, then we have the following cases Once again the case λ 1 = λ 2 = 0 is excluded since then we have that a = b = h = 0 and the equation (41) is meaningless If λ 1 = λ 2 > 0, ie if a = b > 0 and h = 0, then S is a circle centered at the origin with radius 1 a If λ 1 = λ 2 < 0, ie a = b < 0, h = 0, then S = More generally, if λ 1 0 and λ 2 0, we have S = If λ 1 > 0 and λ 2 > 0, then the equation represents an ellipse with centre at the origin The lengths of the semi-axes of the ellipse are 1 λ1 and 1 λ2 The equations of the respective axes are y = 0 and x = 0, which can be written in the original coordinates as v 1 x + v 2 y = 0, u 1 x + u 2 y = 0 If λ 1 > 0 and λ 2 < 0 or if λ 1 < 0 and λ 2 > 0, then we have a hyperbola The lengths and equations of the axes are given as in the case of the ellipse above If λ 1 = λ 2 0, then the equation (42) reduces to x 2 y 2 = 1 λ 1 which is a rectangular hyperbola with axes given by x = ±y In the original coordinates, this reduces to u 1 x + u 2 y = ±(v 1 x + v 2 y) 8

10 5 The general equation of the second degree: straight lines Let us now consider the general equation of the second degree in two variables given by ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 (51) We will try to completely describe the solution set S of this equation Certain computations will repeatedly occur and so it will be useful for us to do them once and for all Set x = X + α and y = Y + β Then (51) becomes ax 2 + 2hXY + by 2 +2(aα + hβ + g)x + 2(hα + bβ + f)y +(aα 2 + 2hαβ + bβ 2 + 2gα + 2fβ + c) = 0 (52) Notice that the constant term in the last line on the left-hand side of the above equation can also be rewritten as We will also set α(aα + hβ + g) + β(hα + bβ + f) + (gα + fβ + c) (53) = a h g h b f g f c We will denote by A the symmetric matrix associated to the quadratic terms, ie [ ] a h A = h b We will use the notations of the previous sections In particular λ 1 and λ 2 will stand for its two eigenvalues An associated pair of normalized eigenvectors will be denoted respectively by (u 1, u 2 ) and (v 1, v 2 ) as before Notice that ab h 2 = det(a) = λ 1 λ 2 Theorem 51 The general equation of the second degree in two variables given by (51) defines a pair of intersecting lines if, and only if, ab h 2 < 0 and = 0 Proof: Let us assume that the equation (51) represents a pair of intersecting lines and that the point of intersection is (α, β) Then, if we set x = X + 10

11 α, y = Y +β, the equation represents a pair of lines intersecting at the origin in the XY -plane Since (α, β) obviously satisfies the equation, we have aα 2 + 2hαβ + bβ 2 + 2gα + 2fβ + c = 0 (54) Further since the equation in the XY variables must be homogeneous, we must have aα + hβ + g = 0, (55) hα + bβ + f = 0, and (52) reduces to ax 2 + 2hXY + by 2 = 0 Since this represents a pair of lines intersecting at the origin, we deduce that ab h 2 < 0 This then implies that the equations (55) admit a unique solution (α, β) which is the point of intersection of these lines in the xyplane Now it follows from (54), the formulation of the left-hand side of this equation given in (53) and the system of equations in (55) that we also have gα + fβ + c = 0 (56) Thus (α, β, 1) is a non-trivial solution to the system of equations aα + hβ + g = 0, hα + bβ + f = 0, gα + fβ + c = 0 (57) This is possible only if = 0 Conversely, if ab h 2 < 0 and = 0, then choose (α, β) as the unique solution of (55) Then, since = 0, it automatically follows that (56) is also satisfied Then, taking into accont (53), we get that (52) reduces to ax 2 + 2hXY + by 2 = 0 under the transformation x = X + α, y = Y + β Now we know that this represents a pair of straight lines intersecting at the origin in the XY -plane and hence (51) represents a pair of straight lines intersecting at (α, β) This completes the proof 11

16 Thus we get a single line x + 2y + 1 = 0 which can be seen to agree with the abstract expression given earlier (iii) Let us take g = 1, c = 1 Then F 2 < λ 2 c In this case, the equation is This can be rewritten as 2x 2 + 8xy + 8y 2 + 2x + 4y + 1 = 0 (x + 2y) 2 + (x + 2y + 1) 2 = 0 It is easy to see that this equation has no solution Thus, all three cases can occur We now summarize the results of this section as follows If = 0 and if ab h 2 < 0, then the equation (51) represents a pair of intersecting straight lines If = 0 and if ab h 2 > 0, then the solution set of the equation (51) consists of a single point If = 0 and if ab h 2 = 0, then the solution set is one of the following: a pair of parallel lines; a single line; the empty set Starting from the coefficients of the equation and the eigenvalues and eigenvectors of the matrix A associated to the quadratic terms, we can determine which of these cases occurs 6 The general equation of the second degree: conics We will consider the equation (51) when 0 Case 1 0 and ab h 2 < 0 Since ab h 2 < 0, there exists a unique solution (α, β) to the system (55) 16

17 However, since 0, we also have that gα + fβ + c 0 Setting x = X + α, y = Y + β, the equation now reads as ax 2 + 2hXY + by 2 + (gα + fβ + c) = 0 Let us set C = gα + fβ + c Then as in Section 3, we can easily see that the solution set is a hyperbola with centre at (α, β) with the lengths of the semi-axes being C and C λ 1 The equations to the axes are given by λ 2 v 1 (x α) + v 2 (y β) = 0, u 1 (x α) + u 2 (y β) = 0, respectively If, in addition, a + b = 0, this will be a rectangular hyperbola Case 2 0 and ab h 2 > 0 Once again, there exists a unique solution (α, β) to the system (55) and setting x = X + α, y = Y + β, the equation reduces to ax 2 + 2hXY + by 2 + C = 0 where C = gα + fβ + c 0 The solution set is either an ellipse or is empty, depending on the signs of C, λ 1 and λ 2 The centre, lengths of the semi-axes and the equations of the axes are exactly as in the previous case Example 61 Consider the equation In this case 5x 2 6xy + 5y x 26y + 29 = 0 = Further ab h 2 > 0 The system (55) reads as 5α 3β + 11 = 0 3α + 5β 13 =

20 If 0 and if ab h 2 < 0, the equation (51) represents a hyperbola If, in addition, a + b = 0, it represents a rectangular hyperbola If 0 and if ab h 2 > 0, then the equation (51) either represents an ellipse or the solution set is empty If a = b and h = 0, then the equation represents a circle provided the solution set is non-empty If 0 and if ab h 2 parabola = 0, then the equation (51) represents a 20

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